A Collapsing Field is indeed Extra Energy from the Ambient Environment

[Debunkified]

Hi everybody. I have an LCR that is exhibiting OU measurements. However True-RMS meters are being used. So I am firmly certain that they can read many different types of waveforms. I do not have an oscilloscope because they are way too much to buy but testing this circuit with an oscilloscope is strongly suggested.

Most believe that a collapsing magnetic field is not free energy, but that's because we are always trying to capture it with a diode. A diode as we all know, or should know, is a kind of load and there is a voltage drop across it. It is this loss of energy through the diode that does not allow us to see anything more than 100%.

Another thing to think about is this. Most of us who have been capturing collapsing magnetic fields for years with diodes may think that the inductive discharge of a coil is an output, but is it really? Why would we ever even begin to think that when we all know full well that the collapsing magnetic field is simply the same energy that came from the power source, so really the collapsing field is not an output but an input. We are better off connecting our loads directly to batteries and powering them with a 100% transfer efficiency without a loss during its transition from the battery and directly to a load. Notice I didn't say electrical efficiency but transfer efficiency.

But here's the kicker, most of the energy from a collapsing field comes from the power supply, but not all of it. I show this by pulsing a resistor and coil in series. The resistor shows the normal 100% transfer of energy from the power supply plus the little bit of extra from the coil when it returns its energy that does not come from the power supply. The load resistor simply has more power on and across it than what the power source is supplying. Notice there is no diode to kill the effect.

In case the schematic doesn't get posted here is a link to that schematic:

https://drive.google.com/open?id=1QW8k2fp-AbmEFXdKQDouOGpJo9wI0Mvy

Also I have a video for those interested right here:

https://youtu.be/yNGRwG9orKU

I have made a few mistakes in my wording so please forgive me of that, it is sometimes difficult for me when making these documentations, but the measurements are right as well as all the values in the schematics.

There's a lot going on with this seemingly simple circuit. For instance if the R5 resistor is removed and a C4 capacitor is connected to the battery in parallel and then the battery is disconnected, then the C3 and C4 capacitors both power R5 load resistor together. They both discharge at the exact same time and at the exact same rate but at all times the C3 voltage remains higher than the C4 voltage as well as its wattage of the R5 load resistor. This is just to show what was immediately happening just before disconnecting the battery and replacing it with a capacitor. The power supply powering the switching is not considered but only the power supply being switched through the load is considered since we only want to observe power going into the load compared to what the power actually is on the load.

Nothing great can be seen or expected if one tries to discharge C4 capacitor and try to charge C3 capacitor with more energy, it's just not happening. It doesn't work like that and works the best when the power supply is a constant and unchanging power supply, such as a battery. However the C3 capacitor does indeed power the R5 load resistor with more watts than the C4 capacitor. So there's more coming out than what's going in.

It would be good to replicate this simple circuit, so please replicate freely and let all of us know what your results were either here or on the YT channel. Since then I have replaced the .01uF C1 with a .02 film capacitor and got slightly better results with a little bit of a higher voltage. It's not hard to replicate and although it's not an extreme amount of excess energy it is excess nevertheless and can be measured with True-RMS voltmeters on the capacitors. The output C3 capacitor is measuring a higher voltage directly in parallel across the R5 load resistor than the input voltage. The input and output current is quite literally identical. Please replicate, I have taken the time and energy to freely open source this simple circuit with expecting nothing in return.

[Debunkified]

Hi everybody. I have an LCR that is exhibiting OU measurements. However True-RMS meters are being used. So I am firmly certain that they can read many different types of waveforms. I do not have an oscilloscope because they are way too much to buy but testing this circuit with an oscilloscope is strongly suggested.

Most believe that a collapsing magnetic field is not free energy, but that's because we are always trying to capture it with a diode. A diode as we all know, or should know, is a kind of load and there is a voltage drop across it. It is this loss of energy through the diode that does not allow us to see anything more than 100%.

Another thing to think about is this. Most of us who have been capturing collapsing magnetic fields for years with diodes may think that the inductive discharge of a coil is an output, but is it really? Why would we ever even begin to think that when we all know full well that the collapsing magnetic field is simply the same energy that came from the power source, so really the collapsing field is not an output but an input. We are better off connecting our loads directly to batteries and powering them with a 100% transfer efficiency without a loss during its transition from the battery and directly to a load. Notice I didn't say electrical efficiency but transfer efficiency.

But here's the kicker, most of the energy from a collapsing field comes from the power supply, but not all of it. I show this by pulsing a resistor and coil in series. The resistor shows the normal 100% transfer of energy from the power supply plus the little bit of extra from the coil when it returns its energy that does not come from the power supply. The load resistor simply has more power on and across it than what the power source is supplying. Notice there is no diode to kill the effect.

In case the schematic doesn't get posted here is a link to that schematic:

https://drive.google.com/open?id=1QW8k2fp-AbmEFXdKQDouOGpJo9wI0Mvy

Also I have a video for those interested right here:

https://youtu.be/yNGRwG9orKU

I have made a few mistakes in my wording so please forgive me of that, it is sometimes difficult for me when making these documentations, but the measurements are right as well as all the values in the schematics.

There's a lot going on with this seemingly simple circuit. For instance if a C4 capacitor is connected to the battery in parallel and then the battery is disconnected, then the C3 and C4 capacitors both power the R5 load resistor together. They both discharge at the exact same time and at the exact same rate but at all times the C3 voltage remains higher than the C4 voltage as well as its wattage of the R5 load resistor. This is just to show what was immediately happening just before disconnecting the battery and replacing it with the C4 capacitor. The power supply powering the switching is not considered but only the power supply being switched through the load is considered since we only want to observe power going into the load compared to what the power actually is on the load.

Nothing great can be seen or expected if one tries to discharge C4 capacitor and try to charge C3 capacitor with more energy, it's just not happening. It doesn't work like that and works the best when the power supply is a constant and unchanging power supply, such as a battery. However the C3 capacitor does indeed power the R5 load resistor with more watts than the C4 capacitor. So we're getting more out than what we put in.

When the R5 load resistor is compared to a direct connection to the power supply, or when the power supply is disconnected, when both the C3 and C4 are discharging through the same R5 load resistor, then we only observe a 100% transfer of energy. When compared through the mosfet switching then we observe something more than 100% on the load. It is the combination of all 100% of the input power combined with the collapsing magnetic field power that causes the power on the load to be greater than the 100% transfer of energy from the power supply alone.

It would be good to replicate this simple circuit, so please replicate freely and let all of us know what your results were either here or on the YT channel. Since then I have replaced the .01uF C1 with a .02 film capacitor and got slightly better results with a little bit of a higher voltage. It's not hard to replicate and although it's not an extreme amount of excess energy it is excess nevertheless and can be measured with True-RMS voltmeters on the capacitors. The output C3 capacitor is measuring a higher voltage directly in parallel across the R5 load resistor than the input voltage, therefore more wattage. The input and output current is quite literally identical. Please replicate, I have taken the time and energy to freely open source this simple circuit and expect nothing in return.

[darediamond]

Whether you use diode or resistor, "A collapsing Field is Indeed Extra Energy From Ambient Enviroment " ONLY AT RESONANT FREQUENCY OF THE INDUCTOR IN THE CIRCUIT.

You can further boost the gained Voltage a and position e resonance current by passing g it via an Farther A.V Pl!ug to suck Negative Electron from The Ground. If you wish, you can further multiply the voltage by passing it via Quadruple Voltage multiplier which is a combination of 2  Avramenko Plugs.

Good luck.

[profitis]

'Most believe that a collapsing magnetic field is not
free energy, but that's because we are always
trying to capture it with a diode. A diode as we
all know, or should know, is a kind of load and
there is a voltage drop across it. It is this loss of
energy through the diode that does not allow us to see anything more than 100%'

whatever thefuck a bakick is i had suggested an almost 100% efficient capture via SERIES nimh battery charging ages ago.you have to know its v before u can do this.